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Current Question (ID: 11236)

Question:
$\text{A particle of mass } m \text{ oscillates with simple harmonic motion between points } x_1 \text{ and } x_2, \text{ the equilibrium position being } O. \text{ Its potential energy is plotted. It will be as given below in the graph:}$
Options:
  • 1. $\text{Parabolic curve with minimum at equilibrium position } O$ (Correct)
  • 2. $\text{Inverted parabolic curve with maximum at equilibrium position } O$
  • 3. $\text{Sinusoidal curve}$
  • 4. $\text{Linear curve}$
Solution:
$\text{Hint: } P.E = \frac{1}{2}Kx^2$ $\text{Explanation: The potential energy of the particle in SHM is given by:}$ $P.E = \frac{1}{2}Kx^2 \text{ [equation of parabola]}$ $\text{At the mean position i.e., } x = 0, \text{ the potential energy of the particle is zero.}$ $\text{At the extreme position i.e., } x = A, \text{ the potential energy of the particle is maximum } P.E = \frac{1}{2}KA^2$ $\text{Therefore, the variation of the potential energy is shown in the figure below:}$ $\text{A parabolic curve with minimum at the equilibrium position } O$ $\text{Hence, option (1) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}